Chapter F, Problem K.14E

(a)

Interpretation Introduction

Interpretation:

Balanced equation for the given skeletal redox reaction has to be written.

    ${\text{Sb}}_{\text{2}}{\text{S}}_{\text{3}}\text{(s)}\text{+}\text{Fe(s)}\to \text{Sb(s)}\text{+}\text{FeS(s)}$

Answer to Problem K.14E

The balanced chemical equation is ${\text{Sb}}_{\text{2}}{\text{S}}_{\text{3}}\text{(s)}\text{+}3\text{Fe(s)}\to 2\text{Sb(s)}\text{+}3\text{FeS(s)}$.

Explanation of Solution

The given skeletal equation for the redox reaction is written as follows;

    ${\text{Sb}}_{\text{2}}{\text{S}}_{\text{3}}\text{(s)}\text{+}\text{Fe(s)}\to \text{Sb(s)}\text{+}\text{FeS(s)}$

Balancing antimony atoms: In the left side of the equation, there are two antimony atoms, while on the right side of the equation; one antimony atom is present.  Adding coefficient $2$ before $\text{Sb}$ balances the antimony atoms on both sides of the equation.  The chemical equation is written as follows;

    ${\text{Sb}}_{\text{2}}{\text{S}}_{\text{3}}\text{(s)}\text{+}\text{Fe(s)}\to 2\text{Sb(s)}\text{+}\text{FeS(s)}$

Balancing sulfur atoms: In the left side of the equation, there are three sulfur atoms, while on the right side of the equation; one sulfur atom is present.  Adding coefficient $3$ before $\text{FeS}$ balances the sulfur atoms on both sides of the equation.  The chemical equation is written as follows;

    ${\text{Sb}}_{\text{2}}{\text{S}}_{\text{3}}\text{(s)}\text{+}\text{Fe(s)}\to 2\text{Sb(s)}\text{+}3\text{FeS(s)}$

Balancing iron atoms: In the left side of the equation, there is one iron atom, while on the right side of the equation; three iron atoms are present.  Adding coefficient $3$ before $\text{Fe}$ balances the iron atoms on both sides of the equation.  The balanced chemical equation is written as follows;

    ${\text{Sb}}_{\text{2}}{\text{S}}_{\text{3}}\text{(s)}\text{+}3\text{Fe(s)}\to 2\text{Sb(s)}\text{+}3\text{FeS(s)}$

(b)

Interpretation Introduction

Interpretation:

Balanced equation for the given skeletal redox reaction has to be written.

    ${\text{BrO}}^{-}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{-}\text{(aq)}\text{+}{\text{Br}}^{-}\text{(aq)}$

Answer to Problem K.14E

The balanced chemical equation is ${\text{3BrO}}^{-}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{-}\text{(aq)}\text{+}2{\text{Br}}^{-}\text{(aq)}$.

Explanation of Solution

The given skeletal equation for the redox reaction is written as follows;

    ${\text{BrO}}^{-}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{-}\text{(aq)}\text{+}{\text{Br}}^{-}\text{(aq)}$

Balancing oxygen atoms: In the left side of the equation, there is one oxygen atom, while on the right side of the equation; three oxygen atoms are present.  Adding coefficient $3$ before ${\text{BrO}}^{-}$ balances the oxygen atoms on both sides of the equation.  The chemical equation is written as follows;

    ${\text{3BrO}}^{-}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{-}\text{(aq)}\text{+}{\text{Br}}^{-}\text{(aq)}$

Balancing bromine atoms: In the left side of the equation, there are three bromine atoms, while on the right side of the equation; two bromine atoms are present.  Adding coefficient $2$ before ${\text{Br}}^{-}$ balances the bromine atoms on both sides of the equation.  The total charge is also balanced in the equation.  The balanced chemical equation is written as follows;

    ${\text{3BrO}}^{-}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{-}\text{(aq)}\text{+}2{\text{Br}}^{-}\text{(aq)}$

(c)

Interpretation Introduction

Interpretation:

Balanced equation for the given skeletal redox reaction has to be written.

    ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{+}\text{C(s)}\to {\text{Cr}}_{\text{3}}{\text{C}}_{\text{2}}\text{(s)}\text{+}\text{CO(g)}$

Answer to Problem K.14E

The balanced chemical equation is ${\text{3Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{+}13\text{C(s)}\to 2{\text{Cr}}_{\text{3}}{\text{C}}_{\text{2}}\text{(s)}\text{+}9\text{CO(g)}$.

Explanation of Solution

The given skeletal equation for the redox reaction is written as follows;

    ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{+}\text{C(s)}\to {\text{Cr}}_{\text{3}}{\text{C}}_{\text{2}}\text{(s)}\text{+}\text{CO(g)}$

Balancing chromium atoms: In the left side of the equation, there are two chromium atoms, while on the right side of the equation; three chromium atoms are present.  Adding coefficient $3$ before ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$ and $2$ before ${\text{Cr}}_{\text{3}}{\text{C}}_{\text{2}}$ balances the chromium atoms on both sides of the equation.  The chemical equation is written as follows;

    ${\text{3Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{+}\text{C(s)}\to 2{\text{Cr}}_{\text{3}}{\text{C}}_{\text{2}}\text{(s)}\text{+}\text{CO(g)}$

Balancing oxygen atoms: In the left side of the equation, there are nine oxygen atoms, while on the right side of the equation; one oxygen atom is present.  Adding coefficient $9$ before $\text{CO}$ balances the oxygen atoms on both sides of the equation.  The chemical equation is written as follows;

    ${\text{3Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{+}\text{C(s)}\to 2{\text{Cr}}_{\text{3}}{\text{C}}_{\text{2}}\text{(s)}\text{+}9\text{CO(g)}$

Balancing carbon atoms: In the left side of the equation, there is one carbon atom, while on the right side of the equation; thirteen carbon atoms are present.  Adding coefficient $13$ before $\text{C}$ balances the carbon atoms on both sides of the equation.  The balanced chemical equation is written as follows;

    ${\text{3Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{+}13\text{C(s)}\to 2{\text{Cr}}_{\text{3}}{\text{C}}_{\text{2}}\text{(s)}\text{+}9\text{CO(g)}$

(d)

Interpretation Introduction

Interpretation:

Balanced equation for the given skeletal redox reaction has to be written.

    $\text{PbS(s)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\to \text{PbO(s)}\text{+}{\text{SO}}_{\text{2}}\text{(g)}$

Answer to Problem K.14E

The balanced chemical equation is $\text{2PbS(s)}\text{+}3{\text{O}}_{\text{2}}\text{(g)}\to 2\text{PbO(s)}\text{+}2{\text{SO}}_{\text{2}}\text{(g)}$.

Explanation of Solution

The given skeletal equation for the redox reaction is written as follows;

    $\text{PbS(s)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\to \text{PbO(s)}\text{+}{\text{SO}}_{\text{2}}\text{(g)}$

Balancing oxygen atoms: In the left side of the equation, there are two oxygen atoms, while on the right side of the equation; three oxygen atoms are present.  Adding coefficient $3$ before ${\text{O}}_{\text{2}}$ and $2$ before $\text{PbO}$, ${\text{SO}}_{\text{2}}$ balances the oxygen atoms on both sides of the equation.  The chemical equation is written as follows;

    $\text{PbS(s)}\text{+}3{\text{O}}_{\text{2}}\text{(g)}\to 2\text{PbO(s)}\text{+}2{\text{SO}}_{\text{2}}\text{(g)}$

Balancing sulfur atoms: In the left side of the equation, there is one sulfur atom, while on the right side of the equation; two sulfur atoms are present.  Adding coefficient $2$ before $\text{PbS}$ balances the sulfur atoms on both sides of the equation.  This step balances the other atoms too.  The balanced chemical equation is written as follows;

    $\text{2PbS(s)}\text{+}3{\text{O}}_{\text{2}}\text{(g)}\to 2\text{PbO(s)}\text{+}2{\text{SO}}_{\text{2}}\text{(g)}$